# QFT: Vacuum invariant, but vacuum correlations aren't

Consider a free scalar field theory. My struggle is that vacuum correlation functions of fields are only Lorentz invariant under a subgroup of Lorentz transformations, despite the invariance of the vacuum under the complete group of Lorentz transformations! I expect that I am making suspect assumptions somewhere.

I expect the vacuum to be invariant under more than just proper, orthochronous Lorentz transformations: I expect the vacuum to be invariant under time reversal invariance and spatial inversion: $$T|0\rangle = |0\rangle$$ and $$P|0\rangle = |0\rangle$$, where these operators act on the field operators as $$T^{-1} \psi(x) T = \psi(\Lambda_Tx)$$ and $$P^{-1} \psi(x) P = \psi(\Lambda_Px)$$ where $$\Lambda_T$$ and $$\Lambda_P$$ are the usual 4x4 matrices for time reversal and inversion.

However, vacuum invariance implies invariance of correlation functions: consider \begin{align*} D(x,y) &= \langle 0| \psi(x) \psi(y) |0\rangle \\ &= \langle 0|P^{-1}P \psi(x) P^{-1}P \psi(y) P^{-1}P|0\rangle \\ &= \langle 0|\psi(\Lambda_Px)\psi(\Lambda_Py)|0\rangle \\ &= D(\Lambda_Px, \Lambda_Py) \end{align*}

This holds similarly for $$T$$, $$D(x,y) = D(\Lambda_Tx,\Lambda_Ty)$$.

However, (see below) I don't think $$D(x,y) = D(\Lambda_Tx,\Lambda_Ty)$$ is true!

The fact that $$D(x,y) = \langle 0| \psi(x) \psi(y) |0\rangle$$ is only invariant ($$D(\Lambda x, \Lambda y) = D(x,y)$$) under proper, orthochronous Lorentz transformations and not generic Lorentz transformations comes up in discussing causality. Invariance under proper, orthochronous transformations means the commutator $$[\psi(x),\psi(y)]$$ will vanish for spacelike $$x-y$$, which it does. Invariance under all transformations would mean the commutator would vanish for timelike $$x-y$$, but it doesn't! See also A question about causality and Quantum Field Theory from improper Lorentz transformation for background.

What am I getting wrong?

My guesses for what's wrong above:

1. The vacuum is not invariant under time reversal and spatial inversion. Seems unlikely to me.
2. The fields transform differently under the operator implementations of $$T$$ and $$P$$. Seems unlikely to me.
3. My insertions of $$I = P^{-1} P$$ and $$I = T^{-1} T$$ are mistaken, perhaps in the latter case by the anti-unitarity of the operator implementation of $$T$$. Unsure.

Are $$C$$, $$P$$ symmetries of your model individually?

If yes, then no wonder your correlation functions are invariant under them.

If no, there doesn’t exist a unitary operator $$P$$ that acts on fields in the way you described. In fact, $$P$$ will act as

$$\psi \rightarrow P \psi P^{\dagger},$$

which when $$P$$ isn’t unitary doesn’t cancel like you expect it to.

W.r.t. $$T$$ – because it is anti-linear, the story is a bit more involved. Unlike with unitary symmetries, anti-unitary symmetries actually don't preserve inner products – they only preserve probabilities. Hence, the correlation function, which is expressed as an inner product, can and will change under time reversal. Its absolute value squared, however, will not (for $$T$$-invariant models; for models with $$T$$ violation, which is the same as $$CP$$ violation due to the $$CPT$$ theorem, it will).

• Thank you. In my particular case, yes, I have $P$ and $T$ invariance of the theory because I had in mind a quadratic scalar theory, so in my case it must boil down to $T$ only preserving absolute values. This makes sense given that in my case $D(x,y)$ is real and nonnegative for spacelike $x-y$ but not necessarily real for timelike $x-y$, showing that $D(x,y) = D(\Lambda_Tx,\Lambda_Ty)$ holds only spacelike $x-y$. Commented Nov 2, 2020 at 20:18